The target reaction contains 2 moles of CO₂ as a product, so we need to multiply equation 1 by two and do the same for the value of ΔH:

1a) 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = -787 kJ

The second equation needs to be reversed since, in the target reaction, CO is a reactant while here it is a product:

2a) 2CO(g) → 2C(s) + O₂(g), ΔH = 221.0 kJ

And now we can add the two equations:

1a) 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = -787 kJ
+
2a) 2CO(g) → 2C(s) + O₂(g), ΔH = 221.0 kJ

2C(s) + 2O₂(g) + 2CO(g) → 2CO₂(g) + 2C(s) + O₂(g)

2CO(g) + O₂(g) → 2CO₂(g)

ΔH = -787 kJ + 221.0 kJ = -566 kJ

The target reaction contains 2 moles of S as a reactant, so we need to multiply equation 1 by two and do the same for the value of ΔH:

1a) 2S(s) + 2O₂(g)  →  2SO₂(g)  ΔH = -594 kJ

The second equation needs to be reversed since in the target reaction, SO₃ is a product while here it is a reactant:

2a) 2SO₂(g) + O₂(g) → 2SO₃(g)  ΔH = -198 kJ

And now we can add the two equations:

1a) 2S(s) + 2O₂(g)  →  2SO₂(g)  ΔH = -594 kJ
+
2a) 2SO₂(g) + O₂(g) → 2SO₃(g)  ΔH = -198 kJ

2S(s) + 2O₂(g) + 2SO₂(g) + O₂(g) → 2SO₂(g) + 2SO₃(g)

2S(s) + 3O₂(g) → 2SO₃(g)

ΔH = -594 kJ + (-198 kJ) = -792 kJ

Because C₄H₈ appears as a reactant in the target reaction, we are not going to reverse equation 2. However, to get rid of C₄H_10, we need to multiply equation 2 by two since there are two moles of C₄H₁₀ in equation 3. Let’s label the new equation as 2a:

2a) 2C₄H₈(g) + 2H₂(g) → 2C₄H₁₀(g), ΔH^o = -252 kJ

 We cannot yet add equations 1, 2a, and 3 because there is going to be H2 only in the left side of the equation, but the target equation does not contain H2. This means we need to also reverse equation 1. Notice that reversing equation 2 for the same reason puts C₄H₈ as product and that is not what we have in the target reaction.

So, let’s reverse equation 1, changing the sign of the enthalpy, and label it as 1a:

1a) 2H₂O(g) → 2H₂(g) + O₂(g), ΔH^o = 571 kJ

Now, we can add equations 1a, 2a, and 3:

1a) 2H₂O(g) → 2H₂(g) + O₂(g), ΔH^o = 571 kJ
+
2a) 2C₄H₈(g) + 2H₂(g) → 2C₄H₁₀(g), ΔH^o = -252 kJ
+
3) 2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l), ΔH^o = -5754 kJ

2H₂O(g) + 2C₄H₈(g) + 2H₂(g) + 2C₄H₁₀(g) + 13O₂(g) → 2H₂(g) + O₂(g) + 2C₄H₁₀(g) + 8CO₂(g) + 10H₂O(l)

2C₄H₈(g) + 12O₂(g) → 8CO₂(g) + 8H₂O(l)

ΔH^o = 571 kJ + (-252 kJ) + (-5754 kJ) = -5435 kJ

The last step is to simplify the equation by dividing it by two. Remember, the same goes for the enthalpy value:

2C₄H₈(g) + 12O₂(g) → 8CO₂(g) + 8H₂O(l), ΔH^o = -2718 kJ

The first two equations seem to be in the correct directions since they contain CO and H₂ as reactants which is what we have in the target equation.

The third equation, however, must be reversed because the methanol here is a reactant while in the target equation, it is a product. So, let’s reverse it and label it as 3a:

3a) 2CO₂(g) + 4H₂O(g) → 2CH₃OH(g) + 3O₂(g), ΔH^o = +1430 kJ

Let’s add equations 1, 2, and 3a and see what we get:

1) 2CO(g) + O₂(g) → 2CO₂(g), ΔH = -566 kJ
+
2) 2H₂(g) + O₂(g) → 2H₂O(g), ΔH^o = -571 kJ
+
3a) 2CO₂(g) + 4H₂O(g) → 2CH₃OH(g) + 3O₂(g), ΔH^o = +1430 kJ

You may notice that there are four moles of water on the left side (equation 3a), but only two moles on the right side (equation 2). This means we need to multiply equation two with its enthalpy value by two. Let’s do that and label the new equation as 2a:

2a) 4H₂(g) + 2O₂(g) → 4H₂O(g), ΔH^o = -1141 kJ

And now, we can add equations 1, 2a, and 3a:

1) 2CO(g) + O₂(g) → 2CO₂(g), ΔH = -566 kJ
+
2a) 4H₂(g) + 2O₂(g) → 4H₂O(g), ΔH^o = -1141 kJ
+
3a) 2CO₂(g) + 4H₂O(g) → 2CH₃OH(g) + 3O₂(g), ΔH^o = +1430 kJ
__________________________________________________

2CO) + O₂ + 4H₂ + 2O₂ + 2CO₂ + 4H₂O → 2CO₂ + 4H₂O + 2CH₃OH + 3O₂

Cancel the same molecules on both side of the equation:

2CO + O₂ + 4H₂ + 2O₂ + 2CO₂ + 4H₂O → 2CO₂ + 4H₂O + 2CH₃OH + 3O₂

ΔH^o = -566 kJ + (-1141 kJ) + 1430 kJ/mol = -277 kJ

So, we have:

2CO(g) + 4H₂(g) → 2CH₃OH(g), ΔH^o = -277 kJ

The last step is to divide the equation by two to match the target reaction:

CO(g) + 2H₂(g) → CH₃OH(g), ΔH^o = -139 kJ

The first reaction needs to be reversed because the CO is a reactant, while in the target reaction it is a product. Let’s reverse it and label it as 1a:

1a) CH₃OH(g) → CO(g) + 2H₂(g), ΔH^o = +139 kJ

The second reaction is good as it is. However, the third reaction needs to be reversed as well in order to make CH₃Cl as a reactant as it is in the target equation:

3a) CH₃Cl(g) + H₂O(g) → CH₃OH(g) + HCl(aq), ΔH^o = +28 kJ

We can now add equations 1a, 2, and 3a:

1a) CH₃OH(g) → CO(g) + 2H₂(g), ΔH^o = +139 kJ
+
2) 2H₂(g) + O₂(g) → 2H₂O(g), ΔH^o = -571 kJ
+
3a) CH₃Cl(g) + H₂O(l) → CH₃OH(g) + HCl(aq), ΔH^o = +28 kJ
____________________________________________

CH₃OH(g) + 2H₂(g) + O₂(g) + CH₃Cl(g) + 2H₂O(l) → CO(g) + 2H₂(g) + 2H₂O(g) + CH₃OH(g) + HCl(aq)

Cancel the same molecules on both side of the equation:

CH₃OH(g) + 2H₂(g) + O₂(g) + CH₃Cl(g) + H₂O(l) → CO(g) + 2H₂(g) + 2H₂O(g) + CH₃OH(g) + HCl(aq)

CH₃Cl(g) + O₂(g) → CO(g) + HCl(g) + H₂O(g)

ΔH^o = +139 kJ + (-571 kJ) + 28 kJ = -404 kJ

The first reaction needs to be reversed because the NH₃ is a product, while in the target reaction it is a reactant. Let’s reverse it and label it as 1a:

1a) 2NH₃(g) → N₂(g) + 3H₂(g), ΔH = +92 kJ

Let’s put the three equations together and see what the sum would look like:

1a) 2NH₃(g) → N₂(g) + 3H₂(g), ΔH = +92 kJ

2) 2H₂(g) + O₂(g) → 2H₂O(g), ΔH = -484 kJ

3) N₂(g) + O₂(g) → 2NO(g), ΔH = 181 kJ

One problem is that the hydrogens in equation 1a and 2 do not cancel each other since we have 3 moles in equation 1a and 2 moles in equation 2. Therefore, let’s multiply equation 1a by two, and multiply equation 2 by three:

1b) 4NH₃(g) → 2N₂(g) + 6H₂(g), ΔH = +184 kJ

2a) 6H₂(g) + 3O₂(g) → 6H₂O(g), ΔH = -1452 kJ

And now let’s put these three equations together:

1b) 4NH₃(g) → 2N₂(g) + 6H₂(g), ΔH = +184 kJ

2a) 6H₂(g) + 3O₂(g) → 6H₂O(g), ΔH = -1452 kJ

3) N₂(g) + O₂(g) → 2NO(g), ΔH = 181 kJ

The problem now is the number of moles of nitrogen in equations 1b and 3. We need to multiply equation 3 by two:

3a) 2N₂(g) + 2O₂(g) → 4NO(g), ΔH = 362 kJ

 And now we can add equations 1b, 2a, and 3a:

1b) 4NH₃(g) → 2N₂(g) + 6H₂(g), ΔH = +184 kJ
+
2a) 6H₂(g) + 3O₂(g) → 6H₂O(g), ΔH = -1452 kJ
+
3a) 2N₂(g) + 2O₂(g) → 4NO(g), ΔH = 362 kJ
____________________________________________

4NH₃(g) + 6H₂(g) + 3O₂(g) + 2N₂(g) + 2O₂(g) → 2N₂(g) + 6H₂(g) + 6H₂O(g) + 2NO(g)

4NH₃(g) + 5O₂(g) → 6H₂O(g) + 2NO(g)

ΔH = +184 kJ + (-1452 kJ) + 362 kJ = -906 kJ

We need to multiply equation 1 by two since there are two moles of CuO(s) in the target equation. Let’s label the new equation as 1a:

1a) 2CuO(s) + 2H₂(g) → 2Cu(s) + 2H₂O(g), ΔH = -170 kJ

Equation 2 seems to be fine, so let’s leave it as it is.

There are 2H₂(g) in equation 1a, and this means that we need to reverse equation 3 and multiply it by two so that we can have 2H₂(g) as a product and cancel them with the ones in equation 1a:

3a) 4HCl(g) → 2H₂(g) + 2Cl₂(g), ΔH = 368 kJ

 Let’s add equations 1a, 2, and 3a and see if they add up:

1a) 2CuO(s) + 2H₂(g) → 2Cu(s) + 2H₂O(g), ΔH = -170 kJ

2) 2Cu(s) + Cl₂(g) → 2CuCl(s), ΔH = -274 kJ

3a) 4HCl(g) → 2H₂(g) + 2Cl₂(g), ΔH = 368 kJ

2CuO(s) + 2H₂(g) + 2Cu(s) + Cl₂(g) + 4HCl(g) → 2Cu(s) + 2H₂O(g) + 2CuCl(s) + 2H₂(g) + 2Cl₂(g)

2CuO(s) + 4HCl(g) → 2CuCl(s) + 2H₂O(g) + Cl₂(g)

ΔH = -170 kJ + (-274 kJ) + 368 kJ = -76 kJ

First, we need to reverse equations (2) and (3) and add equation 1 to them:

1) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)  ΔH  = -891 kJ

+

2a) CO₂(g) → C(s) + O₂(g) ΔH = +393 kJ

+

3a) 2H₂O(l) → 2H₂(g) + O₂(g) ΔH = +572 kJ

___________________________________________

CH₄(g) → C(s) + 2H₂(g)  ΔH = -891 + 393 + 572 = +74 kJ/mol

Finally, reverse the reaction to obtain the enthalpy of formation:

C(s) + 2H₂(g) → CH₄(g) ΔH = -74 kJ/mol